Which of the following environments are low-level areas that can be temporarily or seasonally filled with water?

Rainforest

Swamp

Tundra

Wetland giving brainlist HAS TO BE RIGHT

Answers

Answer 1

Answer:

Rainforest

Explanation:

Swamps are permanently filled with water

and Tundras are dry and cold so the answer should be Rainforest

Answer 2

Answer:

its rainforest it's a it's

Explanation:


Related Questions

A gas with a volume of 4.0 L at a pressure of 0.9 atm is allowed to expand until the pressure drops to 0.20 atm. What is the new volume?

Answers

Answer: 25

Explanation:

The new volume will be 18 L

What is volume?

The volume of a dry gas of a given mass was inversely proportional to its pressure at constant temperature, according to Boyle's law. At moderate pressures and temperatures, most gases behave like ideal gases.

Calculation of new volume is shown as below:

The new volume of can be calculated by using Boyle's law.

[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]

It is given that, [tex]P_{1}[/tex] = 0.9 atm. [tex]V_{1}[/tex] = 4 L, [tex]P_{2}[/tex] = 0.20 atm

Now, put the values of given data in above equation.

[tex]V_{2}[/tex] = [tex]P_{1}V_{1}[/tex] / [tex]P_{2}[/tex].

[tex]V_{2}[/tex] = (0.9) ×4 / 0.20

= 3.6 / 0.20

= 18 L

therefore, the new volume will be 18 L.

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All of the following are symptoms of someone having a dissociative disorder except
A. memory loss
B. increased energy
C. distorted perception
D. blurred sense of identity

Answers

B ! increased energy.

The symptoms of someone having a dissociative disorder are memory loss, distorted perception and a blurred sense of identity except the increased energy.

What is dissociative disorder?

Dissociative disorders are having problems with memory, identifying a person, emotion, perception, behavior changes, and sense of self. It can disrupt mental functioning completely.

Examples of dissociative disorder symptoms are the experience of detachment and loss of memory.

The given examples are all the symptoms of dissociative disorder except the increased energy.

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What is produced when a strong acid reacts with the bicarbonate buffer system in the human body?
O water
O carbonate ions
O bicarbonate ions
O carbonic acid​

Answers

Carbonic acid is produced when a strong acid reacts with the bicarbonate buffer system in the human body.

Answer:

Carbonic acid

Explanation:

The acid reacts with the bicarbonate ion and create carbonic acid

Question 1
Describe a procedure you could follow to determine the specific heat of a 45-g piece of metal.
Instructions

Answers

Answer:

2.1 × 10^-8 J/°c

Explanation:

Is this the answer if yes I would explain if not , that's all I know .

When a 58.8-g piece of hot alloy is placed in 125 g of cold water in a calorimeter, the temperature of the alloy decreases by 106.1°C, while the temperature of the water increases by 10.5°C. The specific heat of the alloy is 0.087 J/(g °C).

Step 1: We know that the heat gained by the water is equal to the heat lost by the alloy. This is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred.

Step 2: We can calculate the heat gained by the water using the formula:

q = mcΔT

where:

q = heat energy

m = mass

c = specific heat

ΔT = change in temperature

For water, the specific heat (c) is 4.18 J/(g °C). So, the heat gained by the water is:

q = (125 g)(4.18 J/(g °C))(10.5 °C) = 5497.5 J

Step 3: The heat lost by the alloy is equal to the heat gained by the water. So, the heat lost by the alloy is also 5497.5 J.

Step 4: We can calculate the specific heat of the alloy using the formula:

c = q/(mΔT)

where:

c = specific heat

q = heat energy

m = mass

ΔT = change in temperature

So, the specific heat of the alloy is:

c = 5497.5 J / (58.8 g)(106.1 °C) = 0.087 J/(g °C)

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William Rankine coined this energy from

A. Mechanical energy
B. Kinetic energy
C. Potential Energy
D. Total energy
E. Chemical Energy

Answers

Answer:

C.Potential Energy

Explanation:

What is 0.738 × 104? Enter your answer in the box.
_____

Answers

The answer would be 76.752 .

HELP! 20 mL of hydrogen measured at 15°C is heated to 35°C.
What is the new volume at the same pressure?

Answers

Answer:either 20.38 or 21.38

Explanation:

i clicked on 10.38 and it was wrong, i also clicked on 22.38 and it was wrong lol

Answer: 21.38 mL

Explanation: i just answered it on ck-12

1. State how increasing the temperature of a gas changes its volume, assuming pressure is held
constant.

Answers

Explanation:

the volume and temperature of a gas have a ditect relationship,as the temperature increases the volume also increases when pressure is held constand, heating the gas increases the kinetic energy of the particles or atoms,causing the gas to expand until the pressure returns to its original value

What is the hybridization of the central atom in CO2? Hybridization = What are the approximate bond angles in this substance? Bond angles = fill in the blank 2

Answers

Answer:

sp

Explanation:

Hybridization is the combination of atomic orbitals to yield equivalent hybrid orbitals of appropriate energy which can participate in bonding.

In every compound there is a central atom. The central atom is usually the least electronegative atom in the molecule. In this case the least electronegative atom in the molecule is carbon.

The bond between carbon and oxygen in CO2 is intermittent between a pure double and a pure triple bond. Hence, carbon is sp hybridized.

(4 pts) Fill in the number of protons and electrons for each product and reactant (two boxes have been filled in for you). (2 pts) Verify that the number of protons on the left side of the chemical equation is equal to the number of protons of the right side. Show your work. (2 pts) Verify that the number of electrons on the left side of the chemical equation is equal to the number of electrons of the right side. Show your work. (3 pts) Which substance is being oxidized

Answers

Answer:

a)

        Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)

#p⁺    30          2              30             2

#e⁻    30          0              28             2

b) 32

c) 30

d) Zn

Explanation:

There is some info missing. I will add the complete question.

Consider the following oxidation/reduction reaction.

         Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)

#p⁺                                                     2

#e⁻                                                     2

a) Fill in the number of protons and electrons for each product and reactant (two boxes have been filled in for you).

b) Verify that the number of protons on the left side of the chemical equation is equal to the number of protons of the right side. Show your work.

c) Verify that the number of electrons on the left side of the chemical equation is equal to the number of electrons of the right side. Show your work.

d) Which substance is being oxidized?

a) The atomic number of Zn is 30 so it will have 30 protons. Since it ts neutral, it will also have 30 electrons. Zn²⁺ will also have 30 protons but it lost 2 electrons, so it has 28 electrons. The atomic number of H⁺ is 1, so each H atom will have 1 proton (2 in total). But since H has 1 electron, and H⁺ lost 1 electron, H⁺ will have 0 electrons. The complete chart is:

        Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)

#p⁺    30          2              30             2

#e⁻    30          0              28             2

b) The total number of protons on the left side is: 30 + 2 = 32.

The total number of protons on the right side is: 30 + 2 = 32.

c) The total number of electrons on the left side is: 30 + 0 = 30.

The total number of electrons on the right side is: 28 + 2 = 30.

d) Zn(s) is oxidized because it loses electrons (from 30 to 28) and its number increases.

A mixture of 65 percent N2 and 35 percent CO2 gases (on a mass basis) enters the nozzle of a turbojet engine at 60 psia and 1400 R with a low velocity, and it expands to a pressure of 12 psia. If the isentropic efficiency of the nozzle is 88 percent, determine:
(a) the exit temperature
(b) the exit velocity of the mixture.
Assume constant specific heats at room temperature.

Answers

Answer:

a. 969.1 R

b. 2237 ft/s

Explanation:

First the apparent specific heats are determined from the mass fractions of the gases:

[tex]c_{p} &=\left(\mathrm{mf} c_{p}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{p}\right) \mathrm{CO}_{2} \\ &=(0.65 \cdot 0.248+0.35 \cdot 0.203) \frac{\mathrm{Btu}}{\mathrm{lbm} \mathrm{R}} \\ &=0.232 \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ c_{v} &=\left(\mathrm{mf} c_{v}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{v}\right)_{\mathrm{CO}_{2}} \\ &=(0.65 \cdot 0.177+0.35 \cdot 0.158) \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ &=0.170 \frac{\mathrm{Btu}}{\mathrm{lbmR}}[/tex]

The isentropic coefficient then is:

[tex]k &=\frac{c_{p}}{c_{v}} \\ &=\frac{0.232}{0.17} \\ &=1.365[/tex]

The final temperature is determined from the isentropic nozzle efficiency relation:

[tex]T_{2} &=T_{1}-\eta_{N}\left(T_{1}-T_{2 s}\right) \\ &=T_{1}\left(1-\eta_{N}\left(1-\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}\right)\right) \\ &=1400\left(1-0.88\left(1-\left(\frac{800}{100}\right)^{(1.365-1) / 1.365}\right)\right) \mathrm{R} \\ &=969.1 \mathrm{R}[/tex]

b. The outlet velocity is determined from the energy balance:

[tex]h_{1} &+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2} \\ v_{2} &=\sqrt{2 c_{p}\left(T_{1}-T_{2}\right)} \\ &=\sqrt{2 \cdot 0.232(1400-969.2) \cdot 25037} \frac{\mathrm{ft}}{\mathrm{s}} \\ &=2237 \frac{\mathrm{ft}}{\mathrm{s}}[/tex]

How many grams of Mn are there in 5.09 moles of Mn?

grams

Answers

Explanation:

n=m/M

m=n*M

m=5,09 mol * 54,9g/mol

m= 279,4g

Solution:

[tex]\text{mass of Mn = 5.09 mol Mn} × \frac{\text{54.94 g Mn}}{\text{1 mol Mn}}[/tex]

[tex]\boxed{\text{mass of Mn = 279.6 g}}[/tex]

In scientific notation, the number 0.00262 is expressed as

Answers

Answer:

2.62*10^-3

Explanation:

To do this by hand, you need to understand the rules of scientific notation.

Answer:

0.00262 in scientific notation is 2.62 x 10^-3

45. Which electron configuration represents an
atom in the excited state?
(1) 1s2 2s2 2p6 3s2
(2) 1s2 2s2 2p6 3s1
(3) 1s2 2s2 2p6 (4) 1s2 2s2 2p5 3s2
(4) 1s 2s2p5352

Answers

Answer:

The correct answer is - (4) 1s2 2s2 2p5 3s2

Explanation:

An excited state is a state when the valence electron has moved to some other higher energy orbital, from its ground state orbital. The ground state has a lower energy level or sublevel. In this case, the higher energy level orbit fills before the lower energy level.

In option 4, the last electron is filled in higher energy orbit 3s2 before filling the lower or ground energy level 2p5, in the ground state it would be 1s2 2s2 2p6 3s1 instead of 1s2 2s2 2p5 3s2.

Thus, the correct answer is option 4.

The question is in the photo​

Answers

Answer:

B

Explanation:

from the diagram.... the population of africa increases so..... ofcourse it exceeds its death rate ....

that's all .... have fun

Complete the following sentence by choosing the best answer:
Waves transfer ____.
a. particles
b. energy
с. light
d. matter

Answers

Answer: it is b give braineless

Explanation:

How many moles of aluminum oxide will be produced from 0.50 mol of oxygen?

4 Al + 3 O2→ 2 Al2O3

Answers

Answer:

[tex]\boxed {\boxed {\sf \frac {1}{3} \ mol \ Al_2O_3 \approx 0.34 \ mol \ Al_2O_3}}[/tex]

Explanation:

We will use stoichiometry to solve this problem. The reaction given has a formula of

[tex]4Al+3O_2 \rightarrow 2Al_2O_3[/tex]

The coefficients tell us the number of moles necessary for the reaction.

The reaction requires 3 moles of oxygen to produce 2 moles of aluminum oxide. We can make a ratio.

[tex]\frac {3 \ mol \ O_2}{ 2\ mol \ Al_2O_3}}[/tex]

Since we are doing the reaction with 0.5 moles of oxygen, we multiply the ratio by that number.

[tex]0.5 \ mol \ O_2 *\frac {3 \ mol \ O_2}{ 2 \ mol \ Al_2O_3}}[/tex]

Flip the ratio so the moles of oxygen cancel each other out.

[tex]0.5 \ mol \ O_2 *\frac {2 \ mol \ Al_2O_3}{ 3 \ mol \ O_2}}[/tex]

[tex]0.5 *\frac {2 \ mol \ Al_2O_3}{ 3 }[/tex]

[tex]\frac {1 \ mol \ Al_2O_3}{ 3 }[/tex]

[tex]\frac {1}{3} \ mol \ Al_2O_3 = 0.33333 \ mol \ Al_2O_3 \approx 0.34 \ mol \ Al_2O_3[/tex]

0.5 moles of oxygen produces 1/3 or approximately 0.34 moles of aluminum oxide.

why is nitrogen used for storage of semen in artificial insemination​

Answers

Answer:

Explanation:

Nitrogen is use to store semen for artificial insemination because if it's inert characteristics which it help to maintain the semen inertness and also it maintain the low temperature of the semen which is good for it's quality.

Nitrogen also help to prevent it's biological deterioration thereby maintaining it's viability and effectiveness.

Write an equation and solve the problem.
3. Dexter buys a package of 38 plates. He already
has 4 plates. He puts an equal number on each of
6 tables. How many plates are on each table?

Answers

Dexter already has 4 and he gets 38 so 4+38=42. Than 42/6=7. 7 plates on each table.

How much energy would be released as 95.0 g of acetic acid crystallized?

Answers

Answer:

-18.54kJ of energy would be released

Explanation:

The enthalpy of crystallization is defined as the heat released when 1 mole of a substance change its state from liquid to solid. The ΔHcrystallization of acetic acid is -11.72kJ/mol (ΔHcrystallization = - ΔHfusion).

To solve this question we must convert the mass of acetic acid to moles in order to find the heat released:

Moles acetic acid -Molar mass: 60.052g/mol-:

95.0g * (1mol / 60.052g) = 1.58 moles

Heat released:

1.58 moles * (-11.72kJ/mol) =

-18.54kJ of energy would be released

6. If 4 grams of salt is dissolved in 79 grams of water, what is the total mass of the final solution? [1 mark]
a) 83g
b)4g
c) 79g
d) 81g

Answers

Answer:

A) 83g

Explanation:

Answer:

mass of solution=mass of solute+mass of solvent

=4+79

. =83g

To what temperature must a balloon, initially at 25°C and 2.00 L, be heated in order to have a volume of 6.00 L

Answers

Answer:

894 deg K

Explanation:

The computation is shown below:

Given that

V1 denotes the initial volume of gas = 2.00 L  

T1 denotes the initial temperature of gas = 25 + 273 = 298 K  

V2 denotes the final volume of gas  = 6.00 L  

T2 = ?

Based on the above information

Here we assume that the pressure is remain constant,

So,  

V1 ÷ T1 = V2 ÷ T2  

T2 = T1 × V2 ÷ V1

= (298)(6) ÷ (2)

= 894 deg K

Comprehension Questions:
1. Changing location
change an
object's mass. (does/does not)
O
does
O does not

Answers

Answer:

It does not change and object's mass

In order to boil water, Jacy places a pan of water on the burner of a stove. By which process of thermal energy transfer does
the burner transfer heat to the pan?
A conduction
B. convection
C. radiation
D. respiration

Answers

Answer:

A campfire is a perfect example of the different kinds of heat transfer. If you boil water in a kettle, the heat is transferred through convection from the fire to the pot.

Explanation:

Americium-241 undergoes radioactive decay to produce neptunium-237. Which particle needs to be added to this equation to show that the total numbers of neutrons and protons are not changed by the reaction?​

Answers

Answer:

alpha particle needs to be added to the equation.

Explanation:

Given;

atomic mass of Americium = 241

atomic mass of neptunium = 237

The difference in atomic mass of the radioactive elements given is calculated as;

241 - 237 = 4

The particle with atomic mass of 4 is alpha particle usually represented with helium atom.

Am²⁴¹ -----> He⁴ + Np²³⁷

Thus, alpha particle needs to be added to the equation.

Answer:

4/2 HE

Explanation:

a p e x!

N 2H 4 + H202 N2 + H 20​

Answers

Answer:

N2 + H20

Explanation:

chemical equation balancer

You have 4.5 x 10^24 particles of C3H8. How many grams are present?

Answers

Answer:

33 g.

Explanation:

Hello there!

In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

[tex]1mol=44.11g=6.022x10^{23}molecules[/tex]

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

[tex]4.5x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} *\frac{44.11g}{1mol}\\\\33g[/tex]

Best regards!

What is the volume of 40.0 grams of argon gas at STP ?

Answers

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic TableMolesSTP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up:                                                                                                       [tex]\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})[/tex][DA] Divide/Multiply [Cancel out units]:                                                         [tex]\displaystyle 24.9235 \ L \ Ar[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

Using 1.8x10-5 for the Ka for acetic acid, along with the weight of sodium acetate and volume of acetic acid used to prepare the buffer, calculate the expected pH of the buffer. Show your work for full credit. How well does it compare to the observed value

Answers

Answer:

pH = 4.95

Explanation:

The buffer was created using 4.0082g sodium acetate, mixed with 10.0 mL of 3.0 M acetic acid and 90 mL of deionized water.

Using H-H equation for acetic buffer:

pH = pKa + log [Acetate] / [Acetic acid]

Where pH is the pH of the buffer

pKa = -log Ka = 4.74

[] Are molar concentration of each specie -The volume of the solution is 100mL = 0.1L-:

[Sodium acetate] = 4.0082g * (1mol / 82.03g) = 0.04886 moles / 0.1L

= 0.4886M

[Acetic acid] = 3.0M * (10mL / 100mL) = 0.300M

pH = 4.74 + log [0.4886M] / [0.300M] =

pH = 4.95

why Bohr's theory was inadequate?​

Answers

Bohr’s theory was inadequate because it did not explain the energies absorbed and released by atoms with more than one electron. It worked well for hydrogen atoms but did not work when applied to more complex atoms.
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